Problem. If $ q(t)$ evolves according to the NLS, then $ r(t)=R[q(t)]$ is supposed to have the simple evolution $ r(t,z) = e^{-iz^2 t}r(0,z)$ (and vice versa). This is what I have been trying to derive.

Attempts. As I understand it, one imposes time evolution on the spectral problem of the form $ \partial_t \psi = M\psi$ . Then the compatibility of the two differential equations for $ \psi$ reduces to $ $ \partial_t Q = \partial_x M + [M,Q+iz\sigma],$ $ where $ [\cdot,\cdot]$ is the commutator. If one seeks $ M$ as a quadratic polynomial in $ z$ then one can recover the NLS (by working down from the $ z^3$ terms and then eventually matching the $ z^0$ terms in the equation above). In particular, I computed $ $ M=-i(z^2+2|q|^2)\sigma -zQ – i\partial_x Q\sigma. $ $

My thought was that imposing this time evolution on $ \psi^\pm$ should lead to time evolution for $ a$ and $ b$ , and hence for $ r$ . To see this, we should find formulas for $ a,b$ in terms of $ \psi^\pm$ . From the relation $ \psi^+ = \psi^- A$ , one finds $ $ \psi^+_1 = a\psi^-_1 + b\psi^-_2, $ $ where $ \psi^+_1$ is the first column of $ \psi^+$ and so on. From this (forming the matrix with $ \psi^-_j$ as the second column and using $ \det \psi^-=1$ ) one can get $ $ a=\det[ \psi^+_1 \ \psi^-_2]\quad\text{and} \quad b = -\det[\psi^+_1 \ \psi^-_1]. $ $

Up to this point, what I have done seems to match various references that I have been following. However, if I now try to use the fact that $ \psi^\pm$ solve the prescribed evolution equation, I seem to get $ a_t=0$ and $ b_t=0$ (and hence $ r_t=0$ ). This is because the matrices $ [\psi^+_1 \ \psi^-_2]$ and $ [\psi^+_1 \ \psi^-_1]$ define solutions to $ \partial_t \psi = M\psi$ (don’t they?) and $ \text{tr}(M)=0$ .

What I was expecting was $ a$ to be constant and $ b(t)=e^{-iz^2t}b(0)$ . So I imagine I have a mistake somewhere, or I am failing to understand something (or many things). Any help or insight (or a reference that would bother with such details) would be much appreciated. Thanks!

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